2e423931c3e01a46976d34476b2f44ec Beasiswa Monbukagakusho Tahun 2013 Di Jepang
Monbukagakusho Physics 2010 Solution
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Transcript of Monbukagakusho Physics 2010 Solution
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
1
Physics
STEP BY STEP SOLUTION
BY : Iskandar Setiadi
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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Solution:
Note that A BP P= . Hence '
' ..
A B
A B A A
F F F F F n FA A A n A
= → = → =
Total volume is always constant. Thus, A BV V↓= ↑ , so ' ' '. . . . .A B A AhA h A h A h n A h hn
= → = → =
The answer is option 5.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
3
Solution:
The pole and the box do not change form, so total force exerted at point A must equal total force exerted at point B. Normal force (N) at point A is in the same direction as AF , so F at point A =
.sin .cos .sinAN F W Fθ θ θ+ = + while F at point B = .cos .sinW Fθ θ− .
Hence .cos . .sin cos sin 1 tan.cos . .sin cos sin 1 tan
A
B
F W m gF W m g
θ θ θ θ θθ θ θ θ θ+ + +
= = =− − −
.
The answer is option 5.
F.sin
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
4
Solution:
Assume that, velocity when the ball is launched = v, we can conclude that xv = constant, while
yv is defined by .( ) . .( )2 2y
T g Tv f t t g g t= = − = −
How could we arrive on this conclusion? On t = 2T , 0yv = (mid-point)
Hence t = 0, 2oyTv g= , while at lim ( ) then
2yt T
Tf t v g→
= − .
The height h is defined by:
2
2
1. . .2
1( . ). . .2 2
1 . . .( )2
oyh v t g t
Th g t g t
h g t T t
= −
= −
= −
.
The answer is option 4.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
5
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
6
Solution:
Let determine the constant ( k∑ ) from two springs
. .k x m g∆ =∑
2.(0, 2 ) (0,5 ).(9,8 / )k m kg m s=∑
24,5 /k N m=∑
When A is pulled downward and released gently:
2 mTk
π=
0,5 2224,5 7
T π π= =
Note that: time elapses from release until A reaches its maximum height, t = 2T
1 0,449
2 7Tt sπ= = =
The nearest/best answer is 0,45
The answer is option 3.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
7
Solution: As mechanic energy is constant, at point P, , 0p m kE E E= = . While at point Q, 0, p k mE E E= = .
Each point at x-axis from graph p kE E+ must be constant. The answer is option 4.
constant ( )
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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m.g
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Solution:
This is similar to Q4. First, we need to determine k (constant) from the lightweight rubber.
. .(0,1) (0,5).(9,8) 49 /exertF m g k k N m= → = → =
When the weight is pulled down by 20 cm (from Fig.2), we’ll have:
2 21 1. .( ) .49.(0,2) 2,2052 2rubberE k x J= ∆ = =
Hence,
rubber ballE E=
2,205 . . (0,5).(9,8).( )J m g h h= =
0,45 45h m cm= =
The answer is option 4.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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Solution:
It is known that 2 31 1
2 2
( ) ( )T RT R
= and 2 lTg
π=
Hence, 2 3
2
2( ) ( )
Rg R
T r
π=
2 3
2
2( ) ( )
Rg R
T r
π=
32
22 ( ) .R R Tg r
π =
32
2 2 .( )R rTg R
π=
2 2 r rTR g
π=
The answer is option 3.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
12
Solution:
Using Black’s principle:
. . . . . .c c w w p pm c T m c T m c T∆ + ∆ = ∆
(100).(0.4).( 20) (200).(4, 2).( 20) (150).(0, 4).(80 )T T T− + − = − 40 800 840 16800 4800 60T T T− + − = − 940 22400T = 23,83T Cο=
The nearest / best answer is 24.
The answer is option 2.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
13
Solution:
1 1
2 2
. . .
. . .PV n R TPV n R T
=
In such a temperature, monoatomic ideal gas at constant pressure and same type of gas can be
written as: 1 1
2 2
(10 ) 57 273 10 12 1,2(10 ) 273
V T x x xV T x
+ += → = → + = −
−
2, 2 2 0,91x x cm= → = . The nearest/best answer is 1.
The answer is option 3.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
14
Solution:
It is obvious that chemical energy from gasoline is converted to kinetic energy of the automobile. While energy which is lost to the environment and cannot be reused heat.
The answer is option 7.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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Solution:
Speed of waves in Earth increases with increasing depth. As the wave-front pattern takes on the same shape as the shoreline, we can infer that wave speed plays a major role in this observation.
The answer is option 2.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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Solution:
In night, ground releases heat faster than sea. Heat from ground is moving through upper sky towards sea, hence, air temperature in the upper sky > air temperature near the ground. Speed of sound (moving particle) is temperature-dependent; as temperature increases the average molecular velocity increases.
The answer is option 4.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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Solution:
The answer is option 4.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
18
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
19
Solution:
V (electrical potential) is defined by qV kr
= .
At point A: V = 2 2 2 2 2 2 2 2
( )(2 ) ( ) (2 ) ( ) (3 ) (2 ) (3 ) (2 )
Q Q Q Qka a a a a a a a+ − + −
+ + ++ + + +
= 0.
At point B: V = 2 2 2 2 2 2 2 2
( )(2 ) (2 ) (2 ) (2 ) (2 ) (2 ) (2 ) (2 )
Q Q Q Qka a a a a a a a+ − + −
+ + ++ + + +
= 0.
At point C: V = 2 2 2 2 2 2 2 2
( )( ) ( ) (3 ) ( ) (3 ) (3 ) (3 ) ( )
Q Q Q Qka a a a a a a a+ − + −
+ + ++ + + +
V = ( ) 02 10 18 10
Q Q Q Qka a a a+ − + −
+ + + >
Therefore, A B CV V V= < .
The answer is option 1.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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Solution:
Fig.1 0 ACdεκ=
Fig.2 0 01 2 0
1 1 1 1 1 2
(2 ) (2 )
dA AC C C A
d d d d
κε ε εκ κ= + = + =
− −
0
2ACd
εκ=
0
2
01
122
AFig d
AFigd
εκ
εκ= = .
The answer is option 2.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
21
Solution:
We can have several approaches here, but first, note that the bulb is connected with 5Ω resistor, so the total current should be less than 2,4 A. From Fig.1, when I = 2A, V = 2V, and R = 1Ω .
.2 6 2figR I A= Ω→ =∑ . Thus the amount of power consumed: 2 2. 2 .1 4P I R W= = = .
The answer is option 2.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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Solution:
First, we know that F = 1 22
.q qkr
while E = 2
qkr
F = E.q
It is known that F = B.q.v
. . .E q B q v=
E vB=
Law of conservation states that
argkinetics ch eE E=
21 .2
mv qV=
2 2. .v qVm
=
2qVvm
=
Since E vB= and 2qVv
m= , thus 2E qV
B m= .
The answer is option 2.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
24
Solution:
Using Maxwell Law, 02
. .4
i d s rdBr
µπ
×=
Blvε =
. .. B l vi R Blv iR
= → =
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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02
0
.( . .(sin . )). .4
r b
r b
I dr d l vr
iR
π µ θ θπ
+
−=∫ ∫
[infinite straight conducting wire]
02
0
.( (sin . ). . ). .4
r b
r b
Id dr l vr
iR
π µθπ
+
−=∫ ∫
[ sin .dθ θ is independent, we can separate it from equation]
02
.(2. . ). .4
r b
r b
I dr l vr
iR
µπ
+
−=∫
0. 1[ ] . .( )2
r br b
I l vriR
µπ
+−−
=
0. 1 1[ ]. .( )2 ( ) ( )
I l vr b r bi
R
µπ
− ++ −=
0. ( ) ( )[ ]. .( )2 ( )( )
I b r b r l vr b r bi
R
µπ
+ + −+ −=
0. ( ) ( )[ ].(2 ).( )2 ( )( )
I b r b r a vr b r bi
R
µπ
+ + −+ −=
02 2
. 2[ ].(2 ).( )2 ( )
I b a vr bi
R
µπ −=
0
2 2
. (2 ).(2 ).( )2
.( )
I b a vi
R r b
µ
π=
−
2 2
2. . . . ..( )
o a b v IiR r bµ
π=
−
The answer is option 2.
Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011
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Solution:
The electromotive force (emf) opposes the change in flux. ddt
ε Φ= −
The answer is option 3.